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Further Mathematics: Post your doubts here!

Thought blocker

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randomcod said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics - Further (9231)/9231_s08_qp_1.pdf
Question 9
Why is the value of lambda assumed to be that?
Thanks!
Click to expand...
9)
If the degree is greater in the numerator oblique asymptotes can occur.
y = (x^2 – 2x + λ)/(x + 1) = (x – 2 + λ/x)/(1 + 1/x)
As x tends to infinity the function gets nearer to y = x – 2 independantly of λ.

A useful observation is that λ = 1 makes the numerator (x – 1)^2
dy/dx = (x^2 + 2x - 2 – λ)/(x + 1)^2
The earlier observation now suggests that we trial is λ = 1 because the numerator in the derivative becomes
x^2 + 2x – 3 = (x – 1)(x + 3) giving dy/dx = 0 at x = 1 and x = -3.
The first value is the one we want, because it gives both slope zero and y = 0.
In other words the x axis is a tangent to C when λ = 1

I leave it to you to graph y = (x^2 – 2x - 4)/(x + 1), but y = 0 when
x^2 – 2x – 4 = 0 leading to x = 1 ± √5
 
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randomcod

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Thought blocker said:
9)
If the degree is greater in the numerator oblique asymptotes can occur.
y = (x^2 – 2x + λ)/(x + 1) = (x – 2 + λ/x)/(1 + 1/x)
As x tends to infinity the function gets nearer to y = x – 2 independantly of λ.

A useful observation is that λ = 1 makes the numerator (x – 1)^2
dy/dx = (x^2 + 2x - 2 – λ)/(x + 1)^2
The earlier observation now suggests that we trial is λ = 1 because the numerator in the derivative becomes
x^2 + 2x – 3 = (x – 1)(x + 3) giving dy/dx = 0 at x = 1 and x = -3.
The first value is the one we want, because it gives both slope zero and y = 0.
In other words the x axis is a tangent to C when λ = 1

I leave it to you to graph y = (x^2 – 2x - 4)/(x + 1), but y = 0 when
x^2 – 2x – 4 = 0 leading to x = 1 ± √5
Click to expand...
Thank you very much for your response.
Sorry for being repetitive, but I am still unsure of why we assume λ = 1
Please please make the description very detailed of why we do this, as I am self-learning and don't know the whole thing.
 
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Help please!!!!!!
Q1:Find the length of the arc of the parabola x=at^2, y=2at, between the points(0,0) and (ap^2,2ap)
Q2: Find the length of the arc of the cycloid x=a(t+sin t), y=a(1-cos t), between the points t=0 and t=pi

Does anybody feel the final integral hard to evaluate?
 
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rexsun said:
Help please!!!!!!
Q1:Find the length of the arc of the parabola x=at^2, y=2at, between the points(0,0) and (ap^2,2ap)
Q2: Find the length of the arc of the cycloid x=a(t+sin t), y=a(1-cos t), between the points t=0 and t=pi

Does anybody feel the final integral hard to evaluate?
Click to expand...
Look at these notes below n try ur self.. :)
The thing is I am busy solving my doubts, I will solve if you dont get. But do refer these notes, and try yourself. :)
i) http://www.nointrigue.com/docs/notes/maths/maths_parametrics.pdf
ii) Coordinate system CH : 3 http://examsolutions.net/maths-revision/syllabuses/Index/period-1/Further-Pure/module.php
All the best if u dont get it thn ask :)
 
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randomcod said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics - Further (9231)/9231_w13_qp_13.pdf
Question 1. Completely lost. Help!
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1 / r(r + 1)(r - 1) = A/r + B/(r + 1) + C/(r - 1) such that the coefficients of terms in r² and r add to zero and the constant comes to 1.

Then the numerator becomes A (r + 1) (r - 1) + B r (r - 1) + C r (r + 1)

= Ar² - A + Br² - Br + Cr² + Cr.

Thus, A + B + C = 0, C - B = 0, and -A = 1. Therefore A = -1, so B + C = 1, and B = C, so B = C = 1/2.

Thus the expression comes to -1/r + 1/2(r + 1) + 1/2(r - 1).

To aid with the summation, it's good to rewrite the above as

1/2 (1/(r + 1) - 1/r) + 1/2 (1/(r - 1) - 1/r).

Then Σ (1/(r + 1) - 1/r) from 2 to n. just comes to -1/2 + 1/(n + 1) (by all the rest cancelling as you can see by writing it out a bit!), and similarly Σ (1/(r - 1) - 1/r) comes to 1/1 - 1/n. From these you can work out the overall formula I think, i.e. half of their sum.

Therefore the answer for when n is infinity, is 1/2 (-1/2 + 1) = 1/4.
 
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